Help understanding a brief proof to show $\mathbb{Z}[X]/(X^2 - 3) \cong \mathbb{Z}[\sqrt{3}]$ I was recently asked to prove that $\mathbb{Z}[X]/(X^2 - 3) \cong \mathbb{Z}[\sqrt{3}] := \{a + \sqrt{3}b\ | a,b \in \mathbb{Z}\}$. I couldn't do it in a limited amount of time, so I received a sketch of the proof but I'm not sure I understand every step. Here they are:
First of all we consider the map $\phi : \mathbb{Z} \to \mathbb{Z}[\sqrt{3}]$ sending $a$ to a $a +0 \sqrt{3}$ and $X$ to $\sqrt{3}$. Then we know by the universal properties of polynomial rings  that $\phi$ is a morphism of rings, and it's the only one such that the following diagram commutes:
$\mathbb{Z} \overset{\phi}{\to} \mathbb{Z}[\sqrt{3}]$ ; $\mathbb{Z} \overset{(*)}{\to} \mathbb{Z}[X]$ and $\mathbb{Z}[X] \overset{\xi}{\to} \mathbb{Z}[\sqrt{3}]$
Then in the same way, according to the universal property of quotients, we can construct a map, say $\psi$, such that the following diagram commutes:
$\mathbb{Z}[X] \overset{\xi}{\to} \mathbb{Z}[\sqrt{3}]$ ; $\mathbb{Z}[X] \overset{\pi}{\to} \mathbb{Z}[X]/\ker(\xi)$ and $\mathbb{Z}[X]/\ker(\xi) \overset{\psi}{\to} \mathbb{Z}[\sqrt{3}]$ where $\pi$ refers to the canonical projection.
Apparently all that was left to show was that $(X^2 - 3) \subseteq \ker(\xi)$ and $im(\xi) = \mathbb{Z}[\sqrt{3}]$.
The major points I didn't get in this proof were:

*

*the definition of our first map $\phi$, I mean why do we send $a$ to $a + 0\sqrt{3}$?

*which map corresponds to (*)?

*how to show that $(X^2 - 3) \subseteq \ker(\xi)$: I would consider $P \in (X^2 - 3)$ and try to show that it also lives in $\ker(\xi)$, however such $P$ would be of the form $Q(X^2 - 3)$ with $Q \in \mathbb{Z}$, but I don't really get anywhere

*why we need to show that $im(\xi) = \mathbb{Z}[\sqrt{3}]$: indeed this means that we need $\xi$ surjective and I don't see why this is necessary (and I didn't find out how to prove it either)

I believe what I ask is not too difficult, but it would really help me to understand each and every step of this proof, in order to be able to do it again myself.
If there's anything useful I forgot to mention for a good understand of this post don't hesitate to tell me.
And by the way, if you want to give me similar isomorphisms of rings so I can get used to those universal properties, don't hesitate either!
Thanks in advance
 A: Departing from the technicalities of your proof, here's an important but of intuition about quotients of polynomial rings:
The ring $R[X]$ is a ring which contains $R$ and one additional element $X$ which fulfills no algebraic relations except those which are required by the ring axioms (for instance, $X(X+1)=X^2+X$ by the distributive property, but $X^2=1$ is not true, because no axiom leads to the square of an arbitrary ring element to be $1$).
The quotient $R/I$ of a ring is a ring with all the features of the ring $R$, but with the additional constraint that all elements of $I$ are replaced by $0$, and whenever a calculation would yield an element of $I$, it yields $0$ instead.
Now the quotient ring $R[X]/(f)$ with any polynomial $f$ means that we first add an element $X$ which follows no rules except the ring axioms. And then we force all elements of $(f)$, that is, all multiples of $f(X)$, to be $0$. In other words, we force $f(X)$ to be $0$. So in total, such a quotient of a polynomial ring means adding an element $X$ which is a root of $f$. But then it's clear that $\mathbb Z[X]/(X^2-3)$ is just $\mathbb Z$ with a root of $X^2-3$ added. $\sqrt 3$ is such a root. So it's no surprise that this ring is isomorphic to $\mathbb Z[\sqrt 3]$.
Your proof is just a technical way to phrase this idea. As is the universal property of polynomial rings: it's basically saying that adding an element to $R$ (resulting in a ring extension) can be rephrased as quotienting the polynomial ring.
A: *

*$\phi:\mathbb{Z}\to\mathbb{Z}[\sqrt{3}]$ does not send $X$ to $\sqrt{3}$. There's no $X$ to send. It's just the inclusion, viewing an integer as an element of $\mathbb{Z}[\sqrt{3}]$ where the coefficient of $\sqrt{3}$ is $0$. It is $\xi: \mathbb{Z}[X]\to\mathbb{Z}[\sqrt{3}]$ that sends $X\mapsto \sqrt{3}$.

*$(*):\mathbb{Z}\to\mathbb{Z}[X]$ is the analogous maps for polynomials. It let's you view an integer as a constant polynomial.

*I actually think this is the easier direction. If $P\in (X^2-3)$ then $P(X)=Q(X)(X^2-3)$ where $Q(X)\in \mathbb{Z}[X]$. Then just plug in $X=\sqrt{3}$ to See that $P(\sqrt{3})=0$ and so $P(X)\in\ker(\xi)$. You should try the other inclusion.

*You could instead show $im(\psi)=\mathbb{Z}[\sqrt{3}]$, but $im(\xi)=im(\psi)$ and $\xi$ is probably a simpler map (no equivalence classes needed) so why not work with it. One way or another if you are going to use the isomorphism theorem which states
$$R/\ker(\phi)\cong im(\phi)$$
to finish the proof then you will have to show the image of the map is the space you want on the right side of the $\cong$.

A few additional notes

*

*I think you want $\pi:\mathbb{Z}[X]\to\mathbb{Z}[X]/\ker(\xi)$ not the other way around.

*The universal property of polynomial rings sounds fancy, but it really just says that if you want to define a homomorphism from a polynomial ring to another ring you only need to 1) define a homomorphism from the coefficients 2) give values to the variables.

